Part A ± Freezing-Point Depression and Boiling-Point Elevation olution made usin

Question

Part A ± Freezing-Point Depression and Boiling-Point Elevation olution made using 833 g of sucrose, C12 H220, in 0.225 kg of water What is the boiling point of a s H20? Express your answer to five significant figures and include the appropriate units The changes in boiling point (16) or freezing point (AT) in degrees Celsius from a pure solvent can be determined from the equations given here, r Hints 105.54°C Submit espectively: moles of solute kilograms of solvent moles of solute-× Kf kilograms of solvent T = ml× kf - My Answers Give Up where m is the molality of the solution, and Kb and Kf are the boiling-point-elevation and freezing-point-depression constants for the solvent, respectively. For water, Correct Kb= 0.512 C-kgsolvent Part B mol solute C kg solvent mol solute What is the freezing point of a solution that contains 27.4 g of urea, CO(NH2)2 in 285 mL water, H2 O? Assume a density of water of 1.00 g/mL Express your answer to three significant figures and include the appropriate units Hints 2.42 Submit My Answers Give Up Incorrect; Try Again; 3 attempts remaining

Explanation / Answer

B)

Apply Colligative properties

This is a typical example of colligative properties.

Recall that a solute ( non volatile ) can make a depression/increase in the freezing/boiling point via:

dTf = -Kf*molality * i

dTb = Kb*molality * i

where:

Kf = freezing point constant for the SOLVENT; Kb = boiling point constant for the SOLVENT;

molality = moles of SOLUTE / kg of SOLVENT

i = vant hoff coefficient, typically the total ion/molecular concentration.

At the end:

Tf mix = Tf solvent - dTf

Tb mix = Tb solvent - dTb

mol of urea = mass/MW = 27.4/60 = 0.45666

now.. kg of water = 0.285 kg

Tf = 0 - 1.86*(0.45666)/0.285 ) = -2.9803 °C

Tf = -2.9803 °C

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